The differential of a function $\displaystyle f(x,y,z)$ is given by

$\displaystyle f_xdx+f_ydy+f_zdz$

To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

$\displaystyle f_x=-2y\cos(xy)\sin(xy)$

$\displaystyle f_y=-2x\cos(xy)\sin(xy)$

$\displaystyle f_z=\cos(z)$

The differential of a function $\displaystyle f(x,y,z)$ is given by

$\displaystyle f_xdx+f_ydy+f_zdz$

To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

$\displaystyle f_x=7y+12xy^2e^{z^2}$

$\displaystyle f_y=7x+18x^2y^2e^{z^2}$

$\displaystyle f_z=12x^2y^3ze^{z^2}$

The total differential of a function $\displaystyle f(x,y,z)$ is given by

$\displaystyle f_xdx+f_ydy+f_zdz$

To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

$\displaystyle f_x=z^4$, $\displaystyle f_y=\sqrt{z}$, $\displaystyle f_z=4xz^3+\frac{y}{2\sqrt{z}}$

The total differential of a function $\displaystyle f(x,y,z)$ is given by

$\displaystyle f_xdx+f_ydy+f_zdz$

To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

$\displaystyle f_x=\frac{8z}{x}$, $\displaystyle f_y=\frac{8z}{y}$, $\displaystyle f_z=8\ln(xy)$

The total differential $\displaystyle dz$ of a function $\displaystyle z=f(x,y)$ is defined as the sum of the partial derivatives of $\displaystyle z$ with respect to each of its variables; that is,

$\displaystyle dz=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy$

In this case, $\displaystyle z=f(x,y)=x^6+3xy-\sin(xy)$, and so we use the sum rule, the rule for derivatives of a variable raised to a power, the rule for the derivative of $\displaystyle \sin(x)$, and the chain rule to calculate the partial derivatives $\displaystyle \frac{\partial f}{\partial x}$ and $\displaystyle \frac{\partial f}{\partial y}$, as shown:

$\displaystyle \frac{\partial f}{\partial x}=6x^5+3y-y\cos(xy)$,

$\displaystyle \frac{\partial f}{\partial y}=3x-x\cos(xy)$.

In both cases, we treated the variable not being differentiated as a constant, and applied the chain rule to $\displaystyle \sin(xy)$ to calculate its partial derivatives. Now that $\displaystyle \frac{\partial f}{\partial x}dx$ and $\displaystyle \frac{\partial f}{\partial y}dy$ have been calculated, all that remains is to substitute them into the definition of the total derivative:

$\displaystyle dz=(6x^5+3y-y\cos(xy))dx+(3x-x\cos(xy))dy$

The total differential $\displaystyle dz$ of a function $\displaystyle z=f(x,y)$ is defined as the sum of the partial derivatives of $\displaystyle z$ with respect to each of its variables; that is,

$\displaystyle dz=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy$

In this case, $\displaystyle z=f(x,y)=xe^x^y$, and so we use the product rule and the rule for differentiating $\displaystyle e^x$ to calculate the partial derivatives $\displaystyle \frac{\partial f}{\partial x}$ and $\displaystyle \frac{\partial f}{\partial y}$, as shown:

$\displaystyle \frac{\partial f}{\partial x}=(1)(e^x^y)+(x)(ye^x^y)=e^x^y+xye^x^y=e^x^y(1+xy)$,

$\displaystyle \frac{\partial f}{\partial y}=(x)xe^x^y=x^2e^x^y$

In both cases, we treated the variable not being differentiated as a constant, and applied the product rule to calculate its partial derivatives. Now that $\displaystyle \frac{\partial f}{\partial x}dx$ and $\displaystyle \frac{\partial f}{\partial y}dy$ have been calculated, all that remains is to substitute them into the definition of the total derivative:

$\displaystyle dz=e^x^y(1+xy)dx+(x^2e^x^y)dy$

What we need to do is take derivatives, and remember the general equation.

$\displaystyle z=e^{x^2+4y^2}\sin(y)$

$\displaystyle w=g(x,y,z)$

$\displaystyle dw=g_xdx+g_ydy+g_zdz$

When taking the derivative with respect to y recall that the product rule needs to be used.

$\displaystyle dz=2xe^{x^2+4y^2}\sin(y)dx+(8y\sin(y)e^{x^2+4y^2}+cos(y)e^{x^2+4y^2}) dy$

What we need to do is take derivatives, and remember the general equation.

$\displaystyle z=e^{x^2+4y^2}\sin(y)$

$\displaystyle w=g(x,y,z)$

$\displaystyle dw=g_xdx+g_ydy+g_zdz$

When taking the derivative with respect to y recall that the product rule needs to be used.

$\displaystyle dz=2xe^{x^2+4y^2}\sin(y)dx+(8y\sin(y)e^{x^2+4y^2}+cos(y)e^{x^2+4y^2}) dy$